Roboguru

Pada reaksi: P(g)+Q(g)→zathasil, diperoleh data sebagai berikut.    Tentukan: a. orde reaksi terhadap P, b. orde reaksi terhadap Q, c. orde reaksi total, d. rumus laju reaksi, serta e. laju reaksi jika [P] dan [Q] masing-masing dinaikkan 3 kali.

Pertanyaan

Pada reaksi: P open parentheses italic g close parentheses and Q open parentheses italic g close parentheses yields zat space hasil, diperoleh data sebagai berikut.


  


Tentukan:

a. orde reaksi terhadap P,

b. orde reaksi terhadap Q,

c. orde reaksi total,

d. rumus laju reaksi, serta

e. laju reaksi jika open square brackets P close square brackets dan open square brackets Q close square brackets masing-masing dinaikkan 3 kali.

Pembahasan Video:

Pembahasan Soal:

Hukum laju reaksi menyatakan sebuah persamaan yang memperlihatkan keterkaitan atau hubungan antara laju reaksi tertentu dengan konsentrasi pereaksinya.

Berdasarkan persamaan reaksi P open parentheses italic g close parentheses and Q open parentheses italic g close parentheses yields zat space hasil, maka persamaan laju reaksi dapat ditulis menjadi:

r double bond k open square brackets P close square brackets to the power of m open square brackets Q close square brackets to the power of n 

dimana m dan n adalah orde masing-masing reaktan.

Penentuan orde reaksi masing-masing reaktan dapat dicari berdasarkan data eksperimen dengan cara membandingkan data laju reaksi dan konsentrasi reaktan yang diketahui. Konsep yang perlu kita gunakan juga adalah hubungan waktu (t) terhadap laju reaksi (r).

r almost equal to 1 over t 

Langkah 1: Tentukan orde X dan orde Y.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell bold Orde bold space P space... space perc point space left parenthesis 1 right parenthesis space dan space left parenthesis 2 right parenthesis end cell row cell k subscript 1 over k subscript 2 open square brackets P subscript 1 close square brackets to the power of m over open square brackets P subscript 2 close square brackets to the power of m open square brackets Q subscript 1 close square brackets to the power of n over open square brackets Q subscript 2 close square brackets to the power of n end cell equals cell r subscript 1 over r subscript 2 end cell row cell k subscript 1 over k subscript 2 open square brackets P subscript 1 close square brackets to the power of m over open square brackets P subscript 2 close square brackets to the power of m open square brackets Q subscript 1 close square brackets to the power of n over open square brackets Q subscript 2 close square brackets to the power of n end cell equals cell fraction numerator 1 over t subscript 1 over denominator 1 over t subscript 2 end fraction end cell row cell open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of m open parentheses fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close parentheses end cell equals cell fraction numerator 1 over 80 over denominator 1 over 40 end fraction end cell row cell open parentheses 1 half close parentheses to the power of m end cell equals cell 40 over 80 end cell row cell open parentheses 1 half close parentheses to the power of m end cell equals cell open parentheses 1 half close parentheses to the power of 1 end cell row m equals 1 row blank blank blank row blank blank cell bold Orde bold space italic Q space... space perc point space left parenthesis 2 right parenthesis space dan space left parenthesis 3 right parenthesis space left parenthesis Substitusi space m equals 1 right parenthesis end cell row cell fraction numerator k subscript 2 over denominator k 3 end fraction open square brackets X subscript 2 close square brackets to the power of m over open square brackets X subscript 3 close square brackets to the power of m open square brackets Y subscript 2 close square brackets to the power of n over open square brackets Y subscript 3 close square brackets to the power of n end cell equals cell r subscript 2 over r subscript 3 end cell row cell fraction numerator k subscript 2 over denominator k 3 end fraction open square brackets P subscript 2 close square brackets to the power of m over open square brackets P subscript 3 close square brackets to the power of m open square brackets Q subscript 2 close square brackets to the power of n over open square brackets Q subscript 3 close square brackets to the power of n end cell equals cell fraction numerator 1 over t subscript 2 over denominator 1 over t subscript 3 end fraction end cell row cell open parentheses fraction numerator 0 comma 2 over denominator 0 comma 4 end fraction close parentheses to the power of 1 open parentheses fraction numerator 0 comma 1 over denominator 0 comma 2 end fraction close parentheses to the power of n end cell equals cell fraction numerator 1 over 40 over denominator 1 fifth end fraction end cell row cell open parentheses 1 half close parentheses open parentheses 1 half close parentheses to the power of n end cell equals cell 5 over 40 end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 5 over 40 division sign 1 half end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 5 over 40 cross times 2 over 1 end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 10 over 40 end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell 1 fourth end cell row cell open parentheses 1 half close parentheses to the power of n end cell equals cell open parentheses 1 half close parentheses squared end cell row n equals 2 end table 

Langkah 2: Tentukan orde total.

table attributes columnalign right center left columnspacing 0px end attributes row cell orde space total end cell equals cell orde space P and orde space Q end cell row cell orde space total end cell equals cell 1 plus 2 end cell row cell orde space total end cell equals 3 end table  

Langkah 3: Tentukan rumus laju reaksi. Substitusi nilai m = 1 dan n = 2 ke dalam rumus laju reaksi.

table attributes columnalign right center left columnspacing 0px end attributes row r equals cell k open square brackets P close square brackets to the power of m open square brackets Q close square brackets to the power of n end cell row r equals cell k open square brackets P close square brackets to the power of 1 open square brackets Q close square brackets squared end cell row r equals cell k open square brackets P close square brackets open square brackets Q close square brackets squared end cell end table    

Langkah 4: Tentukan laju reaksi jika open square brackets P close square brackets dan open square brackets Q close square brackets masing-masing dinaikkan 3 kali.

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets P close square brackets subscript akhir end cell equals cell 3 cross times open square brackets P close square brackets subscript awal end cell row cell open square brackets P close square brackets subscript akhir end cell equals cell 3 open square brackets P close square brackets end cell row cell open square brackets Q close square brackets subscript akhir end cell equals cell 3 cross times open square brackets Q close square brackets subscript awal end cell row cell open square brackets Q close square brackets subscript akhir end cell equals cell 3 open square brackets Q close square brackets end cell row cell laju space akhir space open parentheses r subscript 1 close parentheses end cell equals cell... ? end cell row blank blank blank row cell r subscript 1 over r end cell equals cell fraction numerator k open square brackets P close square brackets subscript akhir open square brackets Q close square brackets subscript akhir over denominator k open square brackets P close square brackets open square brackets Q close square brackets end fraction end cell row cell r subscript 1 over r end cell equals cell fraction numerator k cross times 3 open square brackets X close square brackets cross times 3 open square brackets Y close square brackets over denominator k cross times open square brackets X close square brackets cross times open square brackets Y close square brackets end fraction end cell row cell r subscript 1 over r end cell equals cell fraction numerator 3 cross times 3 over denominator 1 end fraction end cell row cell r subscript 1 over r end cell equals cell 9 over 1 end cell row cell r subscript 1 end cell equals cell 9 cross times r end cell end table   


Dengan demikian, diperoleh hasil bahwa:

a. orde reaksi terhadap P = 1.

b. orde reaksi terhadap Q = 2.

c. orde total = 3.

d. rumus laju reaksi, r double bond k open square brackets P close square brackets open square brackets Q close square brackets squared.

e. laju reaksi jika open square brackets P close square brackets dan open square brackets Q close square brackets masing-masing dinaikkan 3 kali menjadi 9 kali lebih cepat dibandingkan laju awal. 

Jadi, orde P, orde Q, orde total, rumus laju reaksi, dan perubahan laju jika konsentrasi dinaikkan 3 kali berturut-turut adalah 1, 2, 3, r double bond k open square brackets P close square brackets open square brackets Q close square brackets squared, dan 9 kali lebih cepat dibandingkan laju awal. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 07 Oktober 2021

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Gas nitrogen oksida dan gas bromin bereaksi pada 0∘C menurut persamaan reaksi berikut:   2NO(g)+Br2​(g)→2NOBr(g)   Laju reaksi diikuti dengan mengukur pertambahan konsentrasi NOBr dan diperoleh data...

Pembahasan Soal:

a. Orde reaksi terhadap NO, dengan perbandingan data 4 banding data 3 yaitu:
 

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets N O close square brackets to the power of italic x open square brackets Br subscript 2 close square brackets to the power of italic y end cell equals v row cell open square brackets fraction numerator 0 comma 3 over denominator 0 comma 2 end fraction close square brackets to the power of x open square brackets fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close square brackets to the power of y end cell equals cell 108 over 48 end cell row cell open square brackets 3 over 2 close square brackets to the power of x end cell equals cell 9 over 4 end cell row cell open square brackets 3 over 2 close square brackets to the power of x end cell equals cell open parentheses 3 over 2 close parentheses squared end cell row x equals 2 end table
 

b. Orde reaksi terhadap Br subscript 2, dengan membandingkan data 2 dan 1, yaitu:
 

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets N O close square brackets to the power of italic x open square brackets Br subscript 2 close square brackets to the power of italic y end cell equals v row cell open square brackets fraction numerator 0 comma 1 over denominator 0 comma 1 end fraction close square brackets to the power of x open square brackets fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close square brackets to the power of y end cell equals cell 24 over 12 end cell row cell open square brackets 2 close square brackets to the power of x end cell equals cell 2 to the power of 1 end cell row x equals 1 end table
 

c. Persamaan laju reaksi nya yaitu: V double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets

d. Orde reaksi totalnya yaitu: x+y = 2 + 1 = 3

e. Jika menggunakan data nomor 1, maka nilai tetapan jenis(k) yaitu:

V double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets 12 space M forward slash s double bond k open square brackets 0 comma 1 space M close square brackets squared open square brackets 0 comma 1 space M close square brackets k equals fraction numerator 12 space M forward slash s over denominator 0 comma 001 space M cubed end fraction k equals 12 cross times 10 cubed space M to the power of negative sign 2 end exponent middle dot s to the power of negative sign 1 end exponent
 

f. Laju reaksi saat konsentrasi pereaksi masing-masing 0,4 M, yaitu:space

V double bond k open square brackets N O close square brackets squared open square brackets Br subscript 2 close square brackets V equals 12 cross times 10 cubed open square brackets 0 comma 4 close square brackets squared open square brackets 0 comma 4 close square brackets V equals 12 cross times 10 cubed open square brackets 0 comma 16 close square brackets open square brackets 0 comma 4 close square brackets V equals 0 comma 768 cross times 10 cubed V equals 768 space M middle dot s to the power of negative sign 1 end exponent

Jadi orde reaksi, persamaan laju, dan konstata laju seperti diuraikan diatas

0

Roboguru

Pada reaksi: 2X(aq)+Y(aq)→Z(aq) diperoleh data percobaan sebagai berikut:   Tentukan: orde reaksi total, rumus laju reaksi, nilai tetapan laju reaksi (k) dan satuannya, nilai x

Pembahasan Soal:

Salah satu faktor yang mempengaruhi laju reaksi adalah konsentrasi reaktan. Hubungan antara laju reaksi dengan konsentrasi reaktan dirumuskan dalam suatu persamaan laju reaksi.

Reaksi 2 X left parenthesis italic a italic q right parenthesis plus Y left parenthesis italic a italic q right parenthesis yields Z left parenthesis italic a italic q right parenthesis mempunyai persamaan laju reaksi:

italic v equals italic k open square brackets X close square brackets to the power of italic m open square brackets Y close square brackets to the power of italic n 

dengan:

k = tetapan laju reaksi
m = orde (tingkat atau pangkat) reaksi terhadap X
n = orde reaksi terhadap Y

a.   Orde reaksi total
Orde reaksi ditentukan melalui percobaan, tidak berkaitan dengan koefisien reaksi.

  • Menghitung orde reaksi X

    Untuk menghitung orde reaksi X, pilih 2 percobaan dimana Y mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (2).

    italic v subscript 1 over italic v subscript 2 equals fraction numerator italic k open square brackets X close square brackets subscript 1 superscript italic m open square brackets Y close square brackets subscript 1 superscript italic n over denominator italic k open square brackets X close square brackets subscript 2 superscript italic m open square brackets Y close square brackets subscript 2 superscript italic n end fraction fraction numerator 6 comma 0 cross times 10 to the power of negative sign 3 end exponent over denominator 2 comma 4 cross times 10 to the power of negative sign 2 end exponent end fraction equals fraction numerator up diagonal strike italic k begin italic style left parenthesis straight 0 straight comma straight 2 right parenthesis end style to the power of italic m up diagonal strike begin italic style left parenthesis straight 0 straight comma straight 5 right parenthesis end style to the power of italic n end strike over denominator up diagonal strike italic k left parenthesis 0 comma 4 right parenthesis to the power of italic m up diagonal strike italic left parenthesis italic 0 italic comma italic 5 italic right parenthesis to the power of italic n end strike end fraction 1 fourth equals open parentheses 1 half close parentheses to the power of italic m m equals 2 
     
  • Menghitung orde reaksi Y 

    Untuk menghitung orde reaksi Y, pilih 2 percobaan dimana X mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (3).
    italic v subscript 1 over italic v subscript 3 equals fraction numerator italic k open square brackets X close square brackets subscript 1 superscript italic m open square brackets Y close square brackets subscript 1 superscript italic n over denominator italic k open square brackets X close square brackets subscript 3 superscript italic m open square brackets Y close square brackets subscript 3 superscript italic n end fraction fraction numerator 6 comma 0 cross times 10 to the power of negative sign 3 end exponent over denominator 1 comma 20 cross times 10 to the power of negative sign 3 end exponent end fraction equals fraction numerator up diagonal strike italic k italic left parenthesis italic 0 italic comma italic 2 italic right parenthesis to the power of italic m end strike left parenthesis 0 comma 5 right parenthesis to the power of italic n over denominator up diagonal strike italic k italic left parenthesis italic 0 italic comma italic 2 italic right parenthesis to the power of italic m end strike left parenthesis 0 comma 1 right parenthesis to the power of italic n end fraction 5 over 1 equals open parentheses 5 over 1 close parentheses to the power of italic n n equals 1 
     
  • Menghitung orde reaksi total

    Orde space reaksi space total double bond m and n Orde space reaksi space total equals 2 plus 1 Orde space reaksi space total equals 3 


Jadi, orde reaksi totalnya adalah 3.


b.   Rumus laju reaksi

       italic v equals italic k open square brackets X close square brackets to the power of italic m open square brackets Y close square brackets to the power of italic n italic v equals italic k open square brackets X close square brackets squared open square brackets Y close square brackets 


Jadi, rumus laju reaksinya adalah italic v bold equals italic k bold left square bracket italic X bold right square bracket to the power of bold 2 bold open square brackets Y bold close square brackets.


c.   Nilai tetapan laju reaksi 

Misal kita pilih percobaan nomor (1):

        italic v equals italic k open square brackets X close square brackets squared open square brackets Y close square brackets italic k equals fraction numerator italic v over denominator open square brackets X close square brackets squared open square brackets Y close square brackets end fraction italic k equals fraction numerator 6 comma 0 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent space detik to the power of negative sign 1 end exponent over denominator open parentheses 0 comma 2 space mol space L to the power of negative sign 1 end exponent close parentheses squared open parentheses 0 comma 5 space mol space L to the power of negative sign 1 end exponent close parentheses end fraction italic k equals fraction numerator 6 comma 0 cross times 10 to the power of negative sign 3 end exponent space mol space L to the power of negative sign 1 end exponent space detik to the power of negative sign 1 end exponent over denominator 2 comma 0 cross times 10 to the power of negative sign 2 end exponent space mol cubed space L to the power of negative sign 3 end exponent end fraction italic k equals 0 comma 3 space mol to the power of negative sign 2 end exponent space L to the power of 2 space end exponent space detik to the power of negative sign 1 end exponent  


Jadi, nilai tetapan laju reaksinya adalah bold 0 bold comma bold 3 bold space bold mol to the power of bold minus sign bold 2 end exponent bold space italic L to the power of bold 2 bold space end exponent bold space bold detik to the power of bold minus sign bold 1 end exponent.


d.   nilai x

        italic v equals italic k open square brackets X close square brackets squared open square brackets Y close square brackets italic x equals open parentheses 0 comma 3 space mol to the power of negative sign 2 end exponent space L squared space detik to the power of negative sign 1 end exponent close parentheses open parentheses 0 comma 3 space mol space L to the power of negative sign 1 end exponent close parentheses squared open parentheses 0 comma 4 space mol space L to the power of negative sign 1 end exponent close parentheses italic x equals 1 comma 08 cross times 10 to the power of negative sign 2 end exponent space mol space L to the power of negative sign 1 end exponent space detik to the power of negative sign 1 end exponent 


Jadi, nilai x pada percobaan adalah bold 1 bold comma bold 08 bold cross times bold 10 to the power of bold minus sign bold 2 end exponent bold space bold mol bold space italic L to the power of bold minus sign bold 1 end exponent bold space bold detik to the power of bold minus sign bold 1 end exponent.space space space

2

Roboguru

Pada reaksi: 2NO(g)+O2​(g)→N2​O4​(g) diperoleh data sebagai berikut:   Tentukan orde reaksi terhadap masing-masing pereaksi. Tentukan rumus laju reaksinya. Hitung nilai tetapan laju reaksi dan s...

Pembahasan Soal:

Salah satu faktor yang mempengaruhi laju reaksi adalah konsentrasi reaktan. Hubungan antara laju reaksi dengan konsentrasi reaktan dirumuskan dalam suatu persamaan laju reaksi. 

Reaksi 2 N O open parentheses italic g close parentheses and O subscript 2 open parentheses italic g close parentheses yields N subscript 2 O subscript 4 open parentheses italic g close parentheses mempunyai persamaan laju reaksi:

italic v equals italic k open square brackets N O close square brackets to the power of italic x open square brackets O subscript 2 close square brackets to the power of italic y 

dengan:

k = tetapan laju reaksi
x = orde (tingkat atau pangkat) reaksi terhadap N O  
y = orde reaksi terhadap O subscript 2  

Orde reaksi ditentukan melalui percobaan, tidak berkaitan dengan koefisien reaksi.

a.   Orde reaksi terhadap masing-masing pereaksi

  • Orde reaksi N O  

    Untuk menghitung orde reaksi N O, pilih 2 percobaan dimana O subscript 2 mempunyai konsentrasi yang sama, yaitu percobaan (2) dan (3).

    v subscript 2 over v subscript 3 equals fraction numerator italic k open square brackets N O close square brackets subscript 2 superscript italic x open square brackets O subscript 2 close square brackets subscript 2 superscript italic y over denominator italic k open square brackets N O close square brackets subscript 3 superscript italic x open square brackets O subscript 2 close square brackets subscript 3 superscript italic y end fraction fraction numerator 0 comma 02 over denominator 0 comma 08 end fraction equals fraction numerator up diagonal strike italic k open parentheses 0 comma 1 close parentheses to the power of italic x up diagonal strike open parentheses 0 comma 2 close parentheses to the power of italic y end strike over denominator up diagonal strike italic k left parenthesis 0 comma 2 right parenthesis to the power of italic x up diagonal strike left parenthesis 0 comma 2 right parenthesis to the power of italic y end strike end fraction 1 fourth equals open parentheses 1 over 1 close parentheses to the power of italic x italic x equals 2  

     
  • Orde reaksi O subscript bold 2 

    Untuk menghitung orde reaksi O subscript 2, pilih 2 percobaan dimana N O mempunyai konsentrasi yang sama, yaitu percobaan (1) dan (2).

    v subscript 1 over v subscript 2 equals fraction numerator italic k open square brackets N O close square brackets subscript 1 superscript italic x open square brackets O subscript 2 close square brackets subscript 1 superscript italic y over denominator italic k open square brackets N O close square brackets subscript 2 superscript italic x open square brackets O subscript 2 close square brackets subscript 2 superscript italic y end fraction fraction numerator 0 comma 01 over denominator 0 comma 02 end fraction equals fraction numerator up diagonal strike italic k open parentheses 0 comma 1 close parentheses to the power of italic x end strike open parentheses 0 comma 1 close parentheses to the power of italic y over denominator up diagonal strike italic k left parenthesis 0 comma 1 right parenthesis to the power of italic x end strike open parentheses 0 comma 2 close parentheses to the power of italic y end fraction 1 half equals open parentheses 1 half close parentheses to the power of italic y italic y equals 1 


Jadi, orde reaksi terhadap N O bold thin space bold dan bold space O subscript bold 2 berturut-turut adalah 2 dan 1. 


b.   Rumus laju reaksi

      italic v equals italic k open square brackets N O close square brackets to the power of italic x open square brackets O subscript 2 close square brackets to the power of italic y italic v equals italic k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets 


Jadi, persamaan laju reaksinya adalah italic v bold equals italic k begin bold style open square brackets N O close square brackets end style to the power of bold 2 begin bold style open square brackets O subscript 2 close square brackets end style.


c.   Nilai tetapan laju reaksi (k)

      Misal kita ambil percobaan nomor (1)

      italic v equals italic k open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets italic k equals fraction numerator italic v over denominator open square brackets N O close square brackets squared open square brackets O subscript 2 close square brackets end fraction italic k equals fraction numerator 0 comma 01 space M space detik to the power of negative sign 1 end exponent over denominator open parentheses 0 comma 1 space M close parentheses squared open parentheses 0 comma 1 space M close parentheses end fraction space italic k equals 10 space M to the power of negative sign 2 end exponent space detik to the power of negative sign 1 end exponent 


Jadi, nilai tetapan lajunya adalah bold 10 bold space italic M to the power of bold minus sign bold 2 end exponent bold space bold detik to the power of bold minus sign bold 1 end exponent.space space space

0

Roboguru

Berikut ini diberikan data percobaan laju reaksi: Q(g)+2T(g)→T2​Q(g) pada beberapa kondisi.   Jika [Q] dan [T] masing-masing diubah menjadi 0,5 M, harga laju (v) reaksi saat itu adalah ... M/det.

Pembahasan Soal:

Untuk mengetahui laju reaksi tersebut, maka dicari orde reaksi dan konstanta laju reaksi terlebih dulu agar diperoleh persamaan laju reaksinya.

Mencari orde reaksi Q menggunakan data 1 dan 2:

space space space space space space space space space space space space space space space fraction numerator v 1 over denominator v 2 end fraction equals fraction numerator k open square brackets Q close square brackets to the power of x open square brackets T close square brackets to the power of y over denominator k open square brackets Q close square brackets to the power of x open square brackets T close square brackets to the power of y end fraction fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent over denominator 5 cross times 10 to the power of negative sign 2 end exponent end fraction equals fraction numerator k left square bracket 0 comma 1 right square bracket to the power of x left square bracket 0 comma 1 right square bracket to the power of y over denominator k left square bracket 0 comma 2 right square bracket to the power of x left square bracket 0 comma 1 right square bracket to the power of y end fraction space space space space space space space space space space space space space space space space 1 fourth equals open parentheses 1 half close parentheses to the power of x space space space space space space space space space space space space space space space space space space x equals 2 

Mencari orde reaksi T menggunakan data 1 dan 3:

space space space space space space space space space space space space space space space fraction numerator v 1 over denominator v 2 end fraction equals fraction numerator k open square brackets Q close square brackets to the power of x open square brackets T close square brackets to the power of y over denominator k open square brackets Q close square brackets to the power of x open square brackets T close square brackets to the power of y end fraction fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent over denominator 10 to the power of negative sign 4 end exponent end fraction equals fraction numerator k left square bracket 0 comma 1 right square bracket to the power of x left square bracket 0 comma 1 right square bracket to the power of y over denominator k left square bracket 0 comma 1 right square bracket to the power of x left square bracket 0 comma 2 right square bracket to the power of y end fraction space space space space space space space space space space space space space space space space 1 over 8 equals open parentheses 1 half close parentheses to the power of y space space space space space space space space space space space space space space space space space space y equals 3 

Jadi, orde raksi terhadap Q adalah 2 dan orde reaksi terhadap T adalah 3. Sehingga, persamaan laju reaksinya adalah v double bond k open square brackets Q close square brackets squared open square brackets T close square brackets cubed.

Kemudian, mencari nilai konstanta laju reaksi dengan memasukan salah satu data ke persamaan laju reaksi, yaitu sebagai berikut:

space space space space space space space space space space space space space space space space v double bond k open square brackets Q close square brackets squared open square brackets T close square brackets cubed space 1 comma 25 cross times 10 to the power of negative sign 2 end exponent equals k left square bracket 0 comma 1 right square bracket squared left square bracket 0 comma 1 right square bracket cubed space 1 comma 25 cross times 10 to the power of negative sign 2 end exponent equals k point 10 to the power of negative sign 5 end exponent space space space space space space space space space space space space space space space space space k equals fraction numerator 1 comma 25 cross times 10 to the power of negative sign 2 end exponent over denominator 10 to the power of negative sign 5 end exponent end fraction space space space space space space space space space space space space space space space space space space space equals space 1.250   

Laju reaksi untuk [Q] dan [T] adalah 0,5 M, yaitu sebagai berikut:

v double bond k open square brackets Q close square brackets squared left square bracket T cubed right square bracket space space equals 1.250 left parenthesis 0 comma 5 right parenthesis squared left parenthesis 0 comma 5 right parenthesis cubed space space space space space equals 39 comma 0 space M forward slash det  

Jadi, jawaban yang benar adalah E.

0

Roboguru

Berikut ini merupakan data percobaan dari reaksi: A+ B →produk.​ Berdasarkan data tersebut, tentukanlah orde untuk masing-masing pereaksi, hukum laju reaksi, dan nilai tetapan laju reaksinya.

Pembahasan Soal:

*Koreksi soal, open square brackets C close square brackets seharusnya waktu yangspacediperlukan.

Laju reaksi adalah laju pengurangan konsentrasi pereaksi atau penambahan konsentrasi hasil reaksi per satuanspacewaktu.

  • Menghitung orde reaksi A, maka cari open square brackets B close square brackets yang sama (Percobaan 1 danspace2)

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets open square brackets A close square brackets subscript 2 over open square brackets A close square brackets subscript 1 close square brackets to the power of italic x end cell equals cell fraction numerator 1 over t subscript 2 over denominator 1 over t subscript 1 end fraction end cell row cell open square brackets fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close square brackets to the power of italic x end cell equals cell fraction numerator 1 over 40 over denominator 1 over 80 end fraction end cell row cell open square brackets fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close square brackets to the power of x end cell equals cell 80 over 40 end cell row cell open square brackets 2 close square brackets to the power of x end cell equals 2 row x equals 1 end table
 

  • Menghitung orde reaksi B, maka cari open square brackets A close square brackets yang sama (Percobaan 1 danspace3)

table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets open square brackets B close square brackets subscript 3 over open square brackets B close square brackets subscript 1 close square brackets to the power of italic y end cell equals cell fraction numerator 1 over t subscript 3 over denominator 1 over t subscript 1 end fraction end cell row cell open square brackets fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close square brackets to the power of italic y end cell equals cell fraction numerator 1 over 20 over denominator 1 over 80 end fraction end cell row cell open square brackets fraction numerator 0 comma 2 over denominator 0 comma 1 end fraction close square brackets to the power of y end cell equals cell 80 over 20 end cell row cell open square brackets 2 close square brackets to the power of y end cell equals 4 row y equals 4 end table
 

  • Menentukan hukum lajuspacereaksi

table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k space open square brackets A close square brackets to the power of x space open square brackets B close square brackets to the power of y end cell row v equals cell k space open square brackets A close square brackets space open square brackets B close square brackets squared end cell end table
 

  • Menentukan nilai tetapan laju reaksispace(k)
    Substitusi nilai laju dan konsentrasi masing-masing ke salah satu percobaan, misalnya percobaan nomorspace1.

table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k space open square brackets A close square brackets space open square brackets B close square brackets squared end cell row cell 1 over t end cell equals cell k space open square brackets A close square brackets space open square brackets B close square brackets squared end cell row cell 1 over 80 end cell equals cell k space open square brackets 0 comma 1 close square brackets space open square brackets 0 comma 1 close square brackets squared end cell row cell 1 over 80 end cell equals cell k space 0 comma 001 end cell row k equals cell fraction numerator 1 over 80 over denominator 0 comma 001 end fraction end cell row k equals cell 12 comma 5 end cell end table


Jadi,

  • Orde reaksi terhadap A adalah 1,
  • Orde reaksi terhadap B adalah 2,
  • Persamaan laju reaksinya adalah table attributes columnalign right center left columnspacing 0px end attributes row v equals cell k space open square brackets A close square brackets space open square brackets B close square brackets squared end cell end table
  • Nilai tetapan laju reaksinya adalah 12,5.space

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